I need help please help!

A picnic basket was marked up 95% from an original cost of $9.57. Last Friday, Rosanne bought the picnic basket and paid an addi

Montano1993 [528]

Answer:

$20.15

Step-by-step explanation:

Given data

Original cost = $9.57

Markup= 95%

Let us find the new cost of the basket

=95/100*9.57

=0.95*9.57

=$9.0915

Hence the cost of the basket is

= 9.57+9.0915

=18.66

Also, the sales tax is 8%

=8/100*18.66

=0.08*18.66

=1.4928

Therefore, the total amount to be paid is

=18.66+1.4928

=$20.15

8 0

3 years ago

Factorise:<br> 14x^2-10x+12

Oksana_A [137]

Answer:

14x^-10x+12+2=-140x+14

Step-by-step explanation:

Very simple. Just add/multiply the like terms, (14x^-10x & 12+2), and then you'll get the answer, but since 140x and 14 are not like terms, the answer stays as 140x+14.

6 0

3 years ago

Tan39= x/68<br> Plz help. Will give brainliet

lesya [120]

Answer:

$\huge\boxed{x \approx 55}$

Step-by-step explanation:

We need to note that this is an equation, in which we can solve for x.

$tan(39) = \frac{x}{68}$

We can use a calculator to find the tangent of 39 - we type 39 into the calculator then press the "TAN" button to find it's tangent. We find that to be around 0.81.

We now have the formula simplified down to $0.81 = \frac{x}{68}$ . To solve for x, we want to isolate it on one side, so we can multiply both sides by 68 to get rid of the fraction.

We know have $0.81 \cdot 68 = x$

$0.81 \cdot 68 = 55.08 \approx 55$ , so x is approximately 55.

Hope this helped!

6 0

3 years ago

Read 2 more answers

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: